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PICList Thread
'Drive a mosfet transistor with a pic'
1999\02\20@194635 by Marcos Migliorini

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Dear piclisters:

Somebody know if is possible to drive a MOSFET transistor (especifically IRF
540) directly from a pic port pin (16F84)?????

Thanks in advance

Marcos Migliorini

1999\02\20@210609 by Bob Blick

face
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Only if you are lightly loaded and not in a noisy environment. The IRF540
starts to turn on at 4 volts and is not very on at 5 volts. Also you're
really asking for trouble if you have a frisky(high voltage, maybe some
inductance) load. You'll probably end up with a nice modulated oscillator...

Cheers,
Bob


>Somebody know if is possible to drive a MOSFET transistor (especifically IRF
>540) directly from a pic port pin (16F84)?????


http://www.bobblick.com/

1999\02\20@212034 by Reginald Neale

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Marcos asked:

>Somebody know if is possible to drive a MOSFET transistor (especifically IRF
>540) directly from a pic port pin (16F84)?????
>

 There is a variant which is designed for this. Part number is something
 like IRL-540. It's in the Digi-Key catalog. What kind of load are you
 driving? Sometimes you need series resistance to limit the driving
 current into the gate capacitance.

 Reg Neale

1999\02\20@212443 by dave vanhorn

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At 09:43 PM 2/20/99 -0300, Marcos Migliorini wrote:
>Dear piclisters:
>
>Somebody know if is possible to drive a MOSFET transistor (especifically IRF
>540) directly from a pic port pin (16F84)?????
>
>Thanks in advance
>
>Marcos Migliorini


First check your turn-on voltage. I must be less than the guaranteed
minimum output voltage of the pic, worst case supply, temp, etc.

Second, check the gate capacitance, it may take a few uS or more to turn
on. During this interval, the FET will be operating in a linear mode, with
high dissipation. This may or may not be a consideration, it mostly depends
what you are switching, and how often.

1999\02\21@030711 by Quentin

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face
The ones that Reg was refering to is the logic gate Mosfet. In other
words, you put a logic high on the gate and the Mosfet switch on. The
parts I use is the IRLZ24 and IRLZ44.
Go to http://www.irf.com for more info.

Quentin

1999\02\21@053844 by mlsirton

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Hi,

On 20 Feb 99 at 21:43, Marcos Migliorini wrote:
> Somebody know if is possible to drive a MOSFET transistor (especifically IRF
> 540) directly from a pic port pin (16F84)?????

I only have the datasheet for the IRF530 here, I think the IRF540 is
similar (higher current, lower RDS ON).

The IRF530 needs 10V (VGS) to fully open.  Depening on your
application you could use a PIC with 5V or maybe 6.25V and get nice
performance (maybe around 4 Amps for 10VDS, just guessing) but the
MOSFET will run hotter - check the datasheet.

You should use a series resistor between the PIC output and the
MOSFET gate (depending on how fast you want to switch) and possibly a
protection diode if you're switching an inductive load.

There are some MOSFETs made specifically for logic level gate inputs
and these will work happily with a 5V PIC (and maybe even a 3V one).
>From looking at the Digikey catalog IRL540 might be what you want
(and only costs 20 cents more...)

Hope this helps,
Guy - spam_OUTmlsirtonTakeThisOuTspaminter.net.il

1999\02\21@073112 by wwl

picon face
On Sat, 20 Feb 1999 21:43:16 -0300, you wrote:

>Dear piclisters:
>
>Somebody know if is possible to drive a MOSFET transistor (especifically IRF
>540) directly from a pic port pin (16F84)?????
>
>Thanks in advance
>
>Marcos Migliorini
You really need to use a logic-level MOSFET to guarantee turn-on at
5V, but otherwise not much of a problem unless you want very high
switching speed (the PIC is unable to charge the often substantial
gate capacitance quickly). Put 100R or so in series with the gate to
avoid the capacitance causing current spikes in the PIC.

1999\02\22@024531 by Justin Grimm

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I agree the irf540 needs at least 10v to turn on fully, Im using one in a projec
t
now and Im switching12v with a pic to drive the 540. Works well.

Justin

Guy Sirton wrote:

{Quote hidden}

--
JUSTIN GRIMM
Eclipse Energy Systems
Email - eclipsespamKILLspamaccessin.com.au
Manufacturers of high quality Power Inverters
http://www.geocities.com/SiliconValley/Ridge/1839/

1999\02\22@061427 by Marcos Migliorini

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Justin:
Could you plase send me an schematic of your switching circuit?
Thanks
Marcos
----- Original Message -----
From: Justin Grimm <.....eclipseKILLspamspam.....ACCESSIN.COM.AU>
To: <EraseMEPICLISTspam_OUTspamTakeThisOuTMITVMA.MIT.EDU>
Sent: Lunes, 22 de Febrero de 1999 05:03 a.m.
Subject: Re: Drive a mosfet transistor with a pic


>I agree the irf540 needs at least 10v to turn on fully, Im using one in a
project
>now and Im switching12v with a pic to drive the 540. Works well.
>
>Justin
>
>Guy Sirton wrote:
>
>> Hi,
>>
>> On 20 Feb 99 at 21:43, Marcos Migliorini wrote:
>> > Somebody know if is possible to drive a MOSFET transistor
(especifically IRF
{Quote hidden}

1999\02\22@121235 by G.R. Kricorissian

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<DIV>Regarding MOSFET drive with a PIC, Bob Blick suggested:<BR><BR>&gt;... put
a small resistor between the PIC and the mosfet.<BR>&gt;Partly to limit the
current, and partly to limit the current fed back into<BR>&gt;the PIC because of
serious &quot;action&quot; on the drain.</DIV>
<DIV><BR>A good suggestion from Bob, but for an entirely different
reason.&nbsp;&nbsp; A MOSFET is a voltage driven device with a very high input
impedance.&nbsp;&nbsp; Thus, a small series resistor is not going to make any
material difference in the input current, nor in any current &quot;fed
back&quot;.</DIV>
<DIV>&nbsp;</DIV>
<DIV>A small series resistor or ferrite bead is usually inserted in the MOSFET's
gate lead to prevent oscillations during switching.&nbsp; What is does is slow
the rise time because of the timeconstant formed with the MOSFET's gate
capacitance.</DIV>
<DIV>&nbsp;</DIV>
<DIV>The remaining thing is to ensure that the driving source has the required
voltage swing to turn on the MOSFET adequately.&nbsp; As has been mentioned,
there are special MOSFETs which support very low gate drive voltages,
specifically for logic-driven applications.</DIV>
<DIV>&nbsp;</DIV>
<DIV>Hope this is of interest,</DIV>
<DIV>.. Gregg<BR></DIV></BODY></HTML>

</x-html>

1999\02\22@152137 by Bob Blick

face
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Actually what I meant by "action" on the drain of the mosfet was
uncontrolled flyback. If the mosfet is overvoltaged and zeners, having a
little resistance in the gate lead will at least leave the pic surviving
if the mosfet has died.

Of course there are other operational reasons as you mentioned for doing
this, but getting the thing to survive abuse was more the topic I was
trying to address. Good points you made!

Cheers,
Bob

1999\02\22@155046 by Dave VanHorn

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>Actually what I meant by "action" on the drain of the mosfet was
>uncontrolled flyback. If the mosfet is overvoltaged and zeners,
having a
>little resistance in the gate lead will at least leave the pic
surviving
>if the mosfet has died.


That should be fixed, but not this way.

When a mosfet turns off an inductive load, there will be a voltage
spike. If this is within the VDS rating of the FET, then you really
don't have to do any more about it (unless it bothers you for some
other reason like EMI)

Adding series resistance between the PIC and the gate limits the
current that the PIC can supply/remove to the gate capacitance, and
therefore the turnon and turnoff times for the FET are lengthened.
This is one way to limit voltage transients, by allowing the energy in
the inductor to dissipate in the FET during switchoff, but you need a
fast scope to tune it, and you are adding to the power dissipation of
the FET.  A snubber network is another way, but if you're switching
rapidly, then it also dissipates some power on every transition.

In order to see the transients on the drain, you need a fast scope,
and a high voltage FET.
I was fooled in one case by an SMPS with 62V transients on a 70V fet.
No problem right?
I had a 350 MHz DSO, and at no point could I observe them exceeding
62V.  The manufacturer agreed, they should be fine, but the FETs were
blowing up in circuit.  Replacing the 70V part with a 100V part
resulted in 90V spikes. The original FET was punching through before I
could see it on a 350 MHz scope!!

The turn-on oscillation effects are similar, the gate turns on the
FET, and a voltage develops across the source lead inductance, causing
the gate voltage relative to the source to drop, cutting off the FET.
This causes the relative gate voltage to increase, turning on the
FET....... A little series R in the gate lead slows down the turnon,
and avoids that big dI/dT in the source lead.  I've never had a
problem  with this where a PIC was driving the fet, but with 384x
switcher chips that can source an amp into the gate, it's almost a
certanity. I've used about 70-100 ohms in various designs.

1999\02\22@160257 by Bob Blick

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> That should be fixed, but not this way.

Of course not. It will, however, allow for someone(not me) to make a few
mistakes without them being serious.
>
>
> Adding series resistance between the PIC and the gate limits the
> current that the PIC can supply/remove to the gate capacitance, and
> therefore the turnon and turnoff times for the FET are lengthened.
> This is one way to limit voltage transients, by allowing the energy in

That was what I said in my first post... this thread is getting a little
spread out.

Cheers,
Bob

1999\02\22@192203 by A & S

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<DIV><FONT color=#000000 face=Arial size=2>I have been using IRZL-44 MOFETS to
drive fuel injectors directly from a pic. When I wired the PIC's output pin
directly to the MOSFET, it did not work. The injectors did not fire. Putting a
200 ohm resistor in series did the trick. Do not know why, but the IR logic
level MOSFETS have worked fine with the drive from a 16C73A running on 5V. (as
long as there is a resistor)</FONT></DIV>
<DIV><FONT color=#000000 face=Arial size=2></FONT>&nbsp;</DIV>
<DIV><FONT face=Arial size=2>Andy MacDonald</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT>&nbsp;</DIV>
<BLOCKQUOTE
style="BORDER-LEFT: #000000 solid 2px; MARGIN-LEFT: 5px; PADDING-LEFT: 5px">
   <DIV><FONT face=Arial size=2><B>-----Origin</B></FONT></DIV>
   <DIV><FONT face=Arial size=2><B></B></FONT>&nbsp;</DIV>
   <DIV><FONT face=Arial size=2><B>al Message-----</B><BR><B>From: </B>G.R.
   Kricorissian &lt;<A
   href="KILLspamgregg-kKILLspamspamSPYDER-IT.COM">RemoveMEgregg-kTakeThisOuTspamSPYDER-IT.COM</A>&gt;<BR><B>To:
   </B><A href="spamBeGonePICLISTspamBeGonespamMITVMA.MIT.EDU">TakeThisOuTPICLISTEraseMEspamspam_OUTMITVMA.MIT.EDU</A>
   &lt;<A
   href="RemoveMEPICLISTspamTakeThisOuTMITVMA.MIT.EDU">PICLISTEraseMEspam.....MITVMA.MIT.EDU</A>&gt;<BR><B>Date:
   </B>Monday, February 22, 1999 9:12 AM<BR><B>Subject: </B>Re: Drive a mosfet
   transistor with a pic<BR><BR></DIV></FONT>
   <DIV>Regarding MOSFET drive with a PIC, Bob Blick suggested:<BR><BR>&gt;...
   put a small resistor between the PIC and the mosfet.<BR>&gt;Partly to limit
   the current, and partly to limit the current fed back into<BR>&gt;the PIC
   because of serious &quot;action&quot; on the drain.</DIV>
   <DIV><BR>A good suggestion from Bob, but for an entirely different
   reason.&nbsp;&nbsp; A MOSFET is a voltage driven device with a very high
   input impedance.&nbsp;&nbsp; Thus, a small series resistor is not going to
   make any material difference in the input current, nor in any current
   &quot;fed back&quot;.</DIV>
   <DIV>&nbsp;</DIV>
   <DIV>A small series resistor or ferrite bead is usually inserted in the
   MOSFET's gate lead to prevent oscillations during switching.&nbsp; What is
   does is slow the rise time because of the timeconstant formed with the
   MOSFET's gate capacitance.</DIV>
   <DIV>&nbsp;</DIV>
   <DIV>The remaining thing is to ensure that the driving source has the
   required voltage swing to turn on the MOSFET adequately.&nbsp; As has been
   mentioned, there are special MOSFETs which support very low gate drive
   voltages, specifically for logic-driven applications.</DIV>
   <DIV>&nbsp;</DIV>
   <DIV>Hope this is of interest,</DIV>
   <DIV>.. Gregg<BR></DIV></BLOCKQUOTE></BODY></HTML>

</x-html>

1999\02\22@223705 by Dwayne Reid

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Hi there.

Bob and Dave have both made very good points regarding series resistors in
the gate leads of MOSFETs.  I'd just like to add that the resistor should be
located as close as possible to the FET as possible.  I had one engineer
forget that - he kept burning up $20.00 FETs because the resistors were on
the processor board, about 18" away from the FETs.  Moving the resistors
right to the FET gate leads eliminated the problem.

dwayne


>> Adding series resistance between the PIC and the gate limits the
>> current that the PIC can supply/remove to the gate capacitance, and
>> therefore the turnon and turnoff times for the FET are lengthened.


Dwayne Reid   <EraseMEdwaynerspamplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(403) 489-3199 voice          (403) 487-6397 fax

Celebrating 15 years of Engineering Innovation (1984 - 1999)

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Do NOT send unsolicited commercial email to this email address.
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email.

1999\02\22@233441 by Dave VanHorn

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>Bob and Dave have both made very good points regarding series
resistors in
>the gate leads of MOSFETs.  I'd just like to add that the resistor
should be
>located as close as possible to the FET as possible.  I had one
engineer
>forget that - he kept burning up $20.00 FETs because the resistors
were on
>the processor board, about 18" away from the FETs.  Moving the
resistors
>right to the FET gate leads eliminated the problem.


ROTFLMAO!!

I never have that problem, I couldn't fit a trace that long :)
Good point though!  Unlike clock leads, you want the resistor on the
destination end in this case!

1999\02\23@002115 by Marcos Migliorini

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First of all I want to thank you all piclisters very much for your exelent
advises.


I have 2 little  problems:

1) First I have checked my local distribiutors and nobody have logic drive
mosfet transistors :((

2) The power supply of this project are two 9V batteries.

So I need to find a solution, may be driving the mosfet with bipolar
transistor?

Marcos Migliorini
{Original Message removed}

1999\02\23@063521 by Dave VanHorn

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>2) The power supply of this project are two 9V batteries.
>
>So I need to find a solution, may be driving the mosfet with bipolar
>transistor?


You will need a pair of transistors, one to turn it on, and another to
turn it off. Then, another to convert from the pic's 0-5V to 0-18V.
Look at the datasheet for a max712 battery charger for ideas,

1999\02\23@181134 by Sean Breheny

face picon face
Ok, I've got to ask:

Can someone please explain why a resistor is needed and how it could matter
where it was placed?

I could see that the resistor will limit the current which is charging up
the gate,but how this would be likely to hurt either the PIC or the FET I
don't really understand.


Thanks,

Sean



At 11:32 PM 2/22/99 -0500, you wrote:
{Quote hidden}

| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
\--------------=----------------
Save lives, please look at http://www.all.org
Personal page: http://www.people.cornell.edu/pages/shb7
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1999\02\23@182604 by Regulus Berdin

picon face
Sean Breheny wrote:
> Can someone please explain why a resistor is needed and how it could matter
> where it was placed?
>
> I could see that the resistor will limit the current which is charging up
> the gate,but how this would be likely to hurt either the PIC or the FET I
> don't really understand.

It may disturb a read-modify-write instruction in the software.  If the
code pulls it high then low and is read fast, the pin still stay charged
due to gate capacitance and the pic may read it as 1 and the FET may
remain on.  With a resistor in series, this problem is minimized.  The
use of shadow register can eliminate this problem.

regards,
Reggie

1999\02\23@192847 by Joe Lepine

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face
I just was reading David Benson's book "Easy Pic'n" where he mentions the tight
timing requirements of the read-modify-write routine.
He suggests using a NOP to let the pin settle.

Would this eliminate the need for the resistor?
Or is it a current overload that is damaging to the PIC?

Regards, Joe

Regulus Berdin wrote:
{Quote hidden}

1999\02\23@201505 by Sean Breheny

face picon face
Hi Reggie,

Thanks,but the main thing that I don't understand is why this might result
in MOSFETs being damaged,as in the post which I replied to.

I ESPECIALLY don't understand why the positioning of the resistor along the
line between the gate and the PIC would matter when it came to protecting
the FET.

Thanks,

Sean

At 06:26 PM 2/23/99 -0500, you wrote:
>It may disturb a read-modify-write instruction in the software.  If the
>code pulls it high then low and is read fast, the pin still stay charged
>due to gate capacitance and the pic may read it as 1 and the FET may
>remain on.  With a resistor in series, this problem is minimized.  The
>use of shadow register can eliminate this problem.
>
>regards,
>Reggie
>
|
| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
\--------------=----------------
Save lives, please look at http://www.all.org
Personal page: http://www.people.cornell.edu/pages/shb7
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1999\02\23@204012 by Regulus Berdin

picon face
Sean Breheny wrote:
> Thanks,but the main thing that I don't understand is why this might result
> in MOSFETs being damaged,as in the post which I replied to.
The MOSFET will be damaged because it will remain ON always.
Example, if PORTA:0 drives a FET and PORTA:1 is an input, if your code
has:

       movlw   1
       movwf   PORTA           ;turn on FET
       .
       .
       .
       bcf     PORTA,0         ;turn it OFF again for after some time
       nop                     ;this can help if w/ a series resistor
       btfss   PORTA,1         ;test input !!!READ-MODIFY-WRITE
        goto   do_something
       goto    do other

At this point, the pin will remain HIGH because of the btfss reading the
pin again while the gate is still charged.

> I ESPECIALLY don't understand why the positioning of the resistor along the
> line between the gate and the PIC would matter when it came to protecting
> the FET.
I don't understand this also.  Maybe the inductance of the wire will
cancel the capacitive effect??!!

regards,
Reggie

--
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ICQ#:   31651436
URL:    http://www.bigfoot.com/~rberdin

1999\02\23@211043 by Reginald Neale

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Sean asked:

>
>Can someone please explain why a resistor is needed and how it could matter
>where it was placed?
>
>I could see that the resistor will limit the current which is charging up
>the gate,but how this would be likely to hurt either the PIC or the FET I
>don't really understand.
>

 a) PIC can only supply 25mA/pin within rating. Not good to exceed
    even for milliseconds; can screw up other on-chip processes.

 b) real circuits are reactive and contain distributed C, L.

 Reg

1999\02\23@232852 by Mike Keitz

picon face
On Tue, 23 Feb 1999 20:13:13 -0500 Sean Breheny <EraseMEshb7spamspamspamBeGoneCORNELL.EDU>
writes:

>I ESPECIALLY don't understand why the positioning of the resistor
>along the
>line between the gate and the PIC would matter when it came to
>protecting
>the FET.

The resistor spoils the RF gain of the FET.  Unless part of a carefully
designed RF amplifier, RF gain is undesirable because it often leads to
oscillation.  If the FET oscillates, it will heat up and may burn out.

It doesn't matter what chip is driving the FET.  Generally the lower the
impedance of the driver's output, the worse the problem becomes.

The resistor needs to be installed near the FET because a long lead
attached to the gate will have some capacitance to ground and the rest of
the circuit.  The capacitance can couple RF to the FET and cause it to
oscillate.

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1999\02\23@234038 by Tjaart van der Walt

flavicon
face
Sean Breheny wrote:
>
> Ok, I've got to ask:
>
> Can someone please explain why a resistor is needed and how it could matter
> where it was placed?
>
> I could see that the resistor will limit the current which is charging up
> the gate,but how this would be likely to hurt either the PIC or the FET I
> don't really understand.
>
> Thanks,
>
> Sean

The instantaneous current drawn by the gate is very high,
because it is a cap you are charging.

As a rule of thumb, the smaller the cap, the lower the
self-inductance, and the higher the peak current.

If you look in your PIC data sheets, you will find a
well-hidden spec that says you shouldn't connect more than
50pF (or was it 25pF?) *directly* on an output. A MOSFET
can be several hundred (help me here, guys) pF.

Maybe you get away with it, maybe you don't.

--
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1999\02\24@101032 by k3jsch

flavicon
face
I agree.  Usually on some general purpose fets, I'll use between 20-40 ohms on t
he gates.

K. Schauf

On Tuesday, February 23, 1999 10:49 PM, Tjaart van der Walt [SMTP:spamBeGonetjaartSTOPspamspamEraseMEWASP.CO
.ZA] wrote:
: Sean Breheny wrote:
: >
: > Ok, I've got to ask:
: >
: > Can someone please explain why a resistor is needed and how it could matter
: > where it was placed?
: >
: > I could see that the resistor will limit the current which is charging up
: > the gate,but how this would be likely to hurt either the PIC or the FET I
: > don't really understand.
: >
: > Thanks,
: >
: > Sean
:
: The instantaneous current drawn by the gate is very high,
: because it is a cap you are charging.
:
: As a rule of thumb, the smaller the cap, the lower the
: self-inductance, and the higher the peak current.
:
: If you look in your PIC data sheets, you will find a
: well-hidden spec that says you shouldn't connect more than
: 50pF (or was it 25pF?) *directly* on an output. A MOSFET
: can be several hundred (help me here, guys) pF.
:
: Maybe you get away with it, maybe you don't.
:
: --
: Friendly Regards          /"\
:                           \ /
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: KILLspamtjaartspamBeGonespamwasp.co.za  / \ AGAINST HTML MAIL
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1999\02\24@132555 by Marc

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>         bcf     PORTA,0         ;turn it OFF again for after some time
>         nop                     ;this can help if w/ a series resistor
>         btfss   PORTA,1         ;test input !!!READ-MODIFY-WRITE

Never knew that "btfss" does write?! The first, "bcf", _is_ a r-m-w cycle, but t
he
latter IMHO is not!

1999\02\24@165136 by Sean Breheny

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Thanks to all who responded to my questions about the gate resistor.

I guess I should I have known about the maximum capacitance on a PIC pin
spec,but I didn't think that this would damage a PIC. I can see now that
even if it wouldn't damage the PIC, it might prevent it's proper operation.

As for damaging the FET, I was thinking that the gate's high capacitance
would short any RF to ground. However, as was pointed out,if the driving
circuit has a very low impedance,it may be able to introduce RF ringing or
other signals from the long lead into the gate and keep the FET in its
linear region. Thanks for the explanation.

Sean

|
| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
\--------------=----------------
Save lives, please look at http://www.all.org
Personal page: http://www.people.cornell.edu/pages/shb7
@spam@shb7@spam@spamspam_OUTcornell.edu ICQ #: 3329174

1999\02\25@050239 by Justin Grimm

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part 0 2464 bytes
Marcos Migliorini wrote:

> Justin:
> Could you plase send me an schematic of your switching circuit?
> Thanks
> Marcos
> {Original Message removed}

1999\02\25@061657 by Marcos Migliorini

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Justin:

Thank you very much!!!!!
Regards

Marcos

----- Original Message -----
From: Justin Grimm <spamBeGoneeclipsespamKILLspamACCESSIN.COM.AU>
To: <.....PICLISTspam_OUTspamMITVMA.MIT.EDU>
Sent: Jueves, 25 de Febrero de 1999 07:21 a.m.
Subject: Re: Drive a mosfet transistor with a pic


{Quote hidden}

----------------------------------------------------------------------------
----

1999\02\25@082848 by Justin Grimm

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Sorry
By the time I sent it I realised how big it was.

Wolfgang Kynast wrote:

{Quote hidden}

--
JUSTIN GRIMM
Eclipse Energy Systems
Email - TakeThisOuTeclipsespamspamaccessin.com.au
Manufacturers of high quality Power Inverters
http://www.geocities.com/SiliconValley/Ridge/1839/

1999\02\25@132126 by wwl

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On Thu, 25 Feb 1999 18:21:12 +0800, you wrote:

>see attachment
1) don't post binaries to this list - post a link to a webpage or
email only to those interested.
2) Never use JPEG for line-art. This image saved as a 2-colour GIF is
one-TENTH of the size of the file you posted, and  would also come out
clearer due to the lossless nature of the compression.

JPEGS are optimised for PHOTOGRAPHS ONLY and suck big-time for
anything else. Similarly you should never use GIFs for photographs -
JPEG will always be substantially more efficient.



1999\02\26@055013 by Justin Grimm

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Woopsy daisy
Only trying to help

Mike Harrison wrote:

{Quote hidden}

--
JUSTIN GRIMM
Eclipse Energy Systems
Email - eclipseEraseMEspamaccessin.com.au
Manufacturers of high quality Power Inverters
http://www.geocities.com/SiliconValley/Ridge/1839/

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